Ta có: \(x^2+9\ge9\)
\(\Leftrightarrow\left(x^2+9\right)^2\ge81\)
\(\Leftrightarrow-\left(x^2+9\right)^2\le-81\)
\(\Leftrightarrow A=40-\left(x^2+9\right)^2\le-81+40=-41\)
Dấu " = " khi \(x^2=0\Leftrightarrow x=0\)
Vậy \(MAX_A=-41\) khi x = 0
Ta có: x2 \(\ge\) 0 \(\forall\) x
\(\Leftrightarrow\) x2 + 9 \(\ge\) 9 \(\forall\) x
\(\Leftrightarrow\) (x2 + 9)2 \(\ge\) 81 \(\forall\) x
\(\Leftrightarrow\) - (x2 + 9)2 \(\le\) -81 \(\forall\) x
\(\Leftrightarrow\) 40 - (x2 + 9)2 \(\le\) 40 - 81 \(\forall\) x
\(\Leftrightarrow\) A \(\le\) -41 \(\forall\) x
Dấu " = " xảy ra \(\Leftrightarrow\) x2 = 0
\(\Leftrightarrow\) x = 0
Vậy Amax = -41 khi x = 0
Vì \(\left(x^2+9\right)^2\ge0\forall x\)
\(\Rightarrow40-\left(x^2+9\right)^2\ge40\forall x\)
\(\Rightarrow A\ge40\forall x\)
Dấu "=" xảy ra khi \(\left(x^2+9\right)^2=0\)
\(\Rightarrow x^2+9=0\Rightarrow x^2=3^2\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy \(MIN_A=40\Leftrightarrow x=\pm3.\)
cảm ơn 
Mn ai bít lm giúp mk vs cần gấp chiều nay hk r ??
